AP Physics C: Mechanics · Topic 5.1

Topic 5.1: Rotational Kinematics

Unit 5: Torque and Rotational Dynamics10-15% of the multiple-choice section

AP Physics C: Mechanics defines angular velocity as the derivative of angular position and angular acceleration as the derivative of angular velocity. AP Physics 1 defines both as averages. That change is what lets a Physics C question hand you angular position as a function of time.

AP Physics: Unit 5 (topics 5.1 Rotational Kinematics). AP Physics C: Mechanics Unit 5, Topic 5.1. One learning objective, 5.1.A: describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration. Four essential-knowledge statements: 5.1.A.1 defines angular displacement in radians with the relevant equation delta-theta = theta - theta_0, and carries 5.1.A.1.i (a rigid system holds its shape but different points move in different directions, so it cannot be modeled as an object), 5.1.A.1.ii (one direction of angular displacement, clockwise or counterclockwise, is typically indicated as mathematically positive and the other as negative) and 5.1.A.1.iii (a system whose rotation can be well described by the motion of its center of mass may be treated as a single object, with Earth's rotation negligible against its revolution about the Earth and Sun center of mass); 5.1.A.2 defines angular velocity as omega = d-theta/dt; 5.1.A.3 defines angular acceleration as alpha = d-omega/dt; 5.1.A.4 states the analogy with one-dimensional linear motion and carries 5.1.A.4.i (the three constant-angular-acceleration equations, introduced with the condition 'for constant angular acceleration') and 5.1.A.4.ii (graphs of the three quantities against time can be used to find the relationships between them). The two-paragraph boundary statement reads: 'AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of said vectors will not be assessed on the exam.' and 'Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation.' AP Physics 1's Topic 5.1 prints only the second paragraph, worded 'for a point or object'. The two courses differ at 5.1.A.2 and 5.1.A.3 only: AP Physics 1 defines average angular velocity as delta-theta/delta-t and average angular acceleration as delta-omega/delta-t. Suggested skills 1.C, 2.B, 2.D, 3.B; AP Physics 1's Topic 5.1 suggests 1.B, 2.A, 2.D, 3.A, 3.C instead. Printed on the C: Mechanics equation sheet: omega = d-theta/dt, alpha = d-omega/dt and the three constant-angular-acceleration equations. Not printed: delta-theta = theta - theta_0, and the angular integral forms, even though the sheet prints both linear integral forms in its translational column.

What Topic 5.1 requires

Topic 5.1 carries one learning objective, 5.1.A: describe the rotation of a system with respect to time using angular displacement, angular velocity, and angular acceleration. The CED lists four suggested skills for it: 1.C (create qualitative sketches of graphs that represent features of a model or the behavior of the physical system), 2.B (calculate or estimate an unknown quantity with units from known quantities), 2.D (predict new values or factors of change of physical quantities using functional dependence between variables), and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim).

Four essential-knowledge statements sit under 5.1.A.

5.1.A.1 defines angular displacement as the measurement of the angle, in radians, through which a point on a rigid system rotates about a specified axis. Its relevant equation is $\Delta\theta = \theta - \theta_0$ . Three sub-statements follow. 5.1.A.1.i says a rigid system is one that holds its shape but in which different points on the system move in different directions during rotation, and that a rigid system cannot be modeled as an object. 5.1.A.1.ii says one direction of angular displacement about an axis of rotation, clockwise or counterclockwise, is typically indicated as mathematically positive, with the other direction becoming mathematically negative. 5.1.A.1.iii says that if the rotation of a system about an axis may be well described using the motion of the system's center of mass, the system may be treated as a single object, and gives Earth as the example: its rotation about its own axis may be considered negligible when considering its revolution about the center of mass of the Earth and Sun system.

5.1.A.2 says angular velocity is the rate at which angular position changes with respect to time, with the relevant equation

\omega = \frac{d\theta}{dt}

5.1.A.3 says angular acceleration is the rate at which angular velocity changes with respect to time, with the relevant equation

\alpha = \frac{d\omega}{dt}

5.1.A.4 says angular displacement, angular velocity, and angular acceleration around one axis are analogous to linear displacement, velocity, and acceleration in one dimension and demonstrate the same mathematical relationships. Sub-statement 5.1.A.4.i supplies the three constant-angular-acceleration equations, and 5.1.A.4.ii says graphs of the three quantities as functions of time can be used to find the relationships between those quantities.

One word this course drops, and everything that follows from it

AP Physics 1 has a Topic 5.1 with the same title, the same learning-objective code, and a 5.1.A.1 that is word for word identical, sub-statements and all. The two frameworks separate at 5.1.A.2 and 5.1.A.3, and the separation is one word wide.

	AP Physics 1	AP Physics C: Mechanics
5.1.A.2 opens	"Average angular velocity is the average rate at which angular position changes with respect to time."	"Angular velocity is the rate at which angular position changes with respect to time."
its equation	$\omega_{\text{avg}} = \frac{\Delta\theta}{\Delta t}$	$\omega = \frac{d\theta}{dt}$
5.1.A.3 opens	"Average angular acceleration is the average rate at which the angular velocity changes with respect to time."	"Angular acceleration is the rate at which angular velocity changes with respect to time."
its equation	$\alpha_{\text{avg}} = \frac{\Delta\omega}{\Delta t}$	$\alpha = \frac{d\omega}{dt}$

The algebra-based course defines the average and drops the subscript when the motion happens to make the two agree. This course defines the instantaneous value and never needs an average at all.

That is not a notation preference. $\Delta\theta/\Delta t$ is a number you can only compute from the angle at two separate instants, so what the algebra-based framework can ask for is a rate over an interval, or a rate read off a graph. $d\theta/dt$ is a function of time, so a Physics C question can hand you $\theta(t)$ as an algebraic expression and ask what happens at one instant. Worked example 1 below is that question, and there is no way to pose it in the algebra-based course.

The reverse matters too. Because this course works with instantaneous quantities, the average and the instantaneous value can disagree, and the exam can make them disagree on purpose. In worked example 1 the rotor's average angular velocity over the first two seconds is $2.0\ \mathrm{rad/s}$ while its instantaneous angular velocity at the end of that interval is exactly zero.

The boundary statement, both paragraphs

Topic 5.1 carries a boundary statement in two paragraphs. Here it is whole, because the second paragraph is what actually bounds your work and the first is the one students misread:

"AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions. However, the directions of said vectors will not be assessed on the exam."

"Descriptions of the directions of rotational kinematics quantities for a point or rigid body are limited to clockwise and counterclockwise with respect to a given axis of rotation."

AP Physics 1's Topic 5.1 prints only the second of those two paragraphs, and it says "for a point or object" rather than "for a point or rigid body". The first paragraph, the one about manipulating magnitudes using vector conventions, is unique to this course.

Read "vector conventions" carefully. It means signs: you pick a positive sense of rotation, you write every angular quantity with a sign relative to that choice, and you add them algebraically. It does not mean you will be asked which way the angular velocity vector points along the axis. The statement says outright that those directions will not be assessed.

This matters because Topic 5.3 in this same unit prints no boundary statement at all and instead teaches you a right-hand rule for the direction of a torque. The unit is genuinely asymmetric: torque gets a direction in three-dimensional space, and $\omega$ and $\alpha$ get clockwise or counterclockwise. Do not carry the freedom you are given in 5.3 backwards into 5.1.

When the three constant-alpha equations are legal, and when they are not

Statement 5.1.A.4.i gives the three equations, and it opens with a condition that is the whole point of the sentence: "For constant angular acceleration, the mathematical relationships between angular displacement, angular velocity, and angular acceleration can be described with the following equations."

\omega = \omega_0 + \alpha t

\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2

\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)

All three are printed on the AP Physics C: Mechanics equation sheet, and all three are printed on the AP Physics 1 sheet in exactly the same form. Nothing separates the two courses here.

What separates them is that this course can give you an $\alpha$ that is not constant, at which point every one of those three equations is simply false. There is no version of them with an average angular acceleration substituted in. Worked example 2 shows what you get if you try: with $\alpha(t) = ct^2$ , using the instantaneous $\alpha$ at the final moment in $\theta = \frac{1}{2}\alpha t^2$ overstates the angle by a factor of six.

The test for whether you may use the kit is not whether the problem mentions angular acceleration. It is whether $\alpha$ is a constant. Three signals that it is not:

The problem gives you $\theta(t)$ or $\omega(t)$ as an expression with a $t^3$ or higher power in it.
The problem gives you a torque that changes with time or with angle, since $\alpha = \tau_{\text{net}}/I$ inherits that change.
The problem gives you a graph of $\alpha$ against $t$ that is not a horizontal line.

In all three cases you go back to $\omega = d\theta/dt$ and $\alpha = d\omega/dt$ , and you differentiate or integrate.

Where the three equations come from, and why deriving them is a graded skill

The constant-angular-acceleration equations are not axioms. They are what you get by integrating the two definitions once each, and this course expects you to be able to do that.

Start from 5.1.A.3 with $\alpha$ constant. Then $d\omega = \alpha\, dt$ , and integrating both sides from time zero to time $t$ gives $\omega - \omega_0 = \alpha t$ , which is the first equation. Substitute that into 5.1.A.2, so $d\theta = (\omega_0 + \alpha t)\, dt$ , and integrate again to get $\theta - \theta_0 = \omega_0 t + \frac{1}{2}\alpha t^2$ , which is the second. Eliminate $t$ between them and you have the third. The whole derivation is four lines and uses only the power rule, which the equation sheet's calculus box prints for you.

It is worth doing once by hand because skill 2.A, deriving a symbolic expression from known quantities by selecting and following a logical mathematical pathway, is weighted at 25 to 30% of the multiple-choice section of this exam. That is the highest band the CED gives any single skill on that section; the next is 2.B at 20 to 25%. The CED's Unit 5 page says students in this unit "will be introduced to new, but somewhat familiar, equations" and will be expected to derive new expressions from them, just as in previous units.

The derivation also tells you exactly which step fails when $\alpha$ is not constant: you cannot pull $\alpha$ out of the first integral. That is the whole failure, and it is why worked example 2's numbers come out where they do.

What the equation sheet prints, and the two integrals it leaves out

The rotational column of the C: Mechanics equation sheet prints five Topic 5.1 equations and omits one.

Equation	CED statement	On the sheet
$\Delta\theta = \theta - \theta_0$	5.1.A.1	no
$\omega = \frac{d\theta}{dt}$	5.1.A.2	yes
$\alpha = \frac{d\omega}{dt}$	5.1.A.3	yes
$\omega = \omega_0 + \alpha t$	5.1.A.4.i	yes
$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$	5.1.A.4.i	yes
$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$	5.1.A.4.i	yes

The two derivative definitions are the rows AP Physics 1 does not have. Its sheet prints the three constant-angular-acceleration equations and nothing above them.

There is one more asymmetry worth knowing before the exam, and it is inside the C sheet rather than between the two sheets. The translational column prints both integral forms, $\Delta x = \int v_x(t)\, dt$ and $\Delta v_x = \int a_x(t)\, dt$ . The rotational column prints neither of their angular analogues. So when you need

\Delta\omega = \int \alpha(t)\, dt \qquad \text{and} \qquad \Delta\theta = \int \omega(t)\, dt

you write them yourself. Statement 5.1.A.4 licenses that: it says the angular quantities "demonstrate the same mathematical relationships" as their linear counterparts. The relationships transfer; the printed lines do not.

One further caution about the sheet, since it is the kind of thing that gets checked wrongly. The relation $s = r\theta$ is printed, but in the geometry box rather than the mechanics table, and it is printed on the AP Physics 1 sheet too. It belongs to Topic 5.2 in both courses.

Reading angular graphs the way this course reads them

Statement 5.1.A.4.ii says graphs of angular displacement, angular velocity, and angular acceleration as functions of time can be used to find the relationships between those quantities. Skill 1.C, creating qualitative sketches of graphs, is one of the four suggested skills for this topic.

The two operations are the two definitions read backwards.

The slope of a $\theta$ against $t$ graph at an instant is $\omega$ at that instant, because $\omega = d\theta/dt$ . The slope of an $\omega$ against $t$ graph is $\alpha$ .
The area under an $\omega$ against $t$ graph between two times is $\Delta\theta$ , and the area under an $\alpha$ against $t$ graph is $\Delta\omega$ . Area below the axis counts as negative.

AP Physics 1 reads graphs the same way, so this section is not a difference between the courses. What is different is that the slope you are asked for here can be the slope of a curve rather than of a line segment, and the area can be the area under a curve rather than under a run of rectangles and triangles. When the shape is a curve, the calculus box on the sheet supplies the rules you need for the integral.

Two sketching habits that pay off on a translation-between-representations question:

A maximum or minimum of $\theta(t)$ sits where $\omega$ crosses zero, and a maximum or minimum of $\omega(t)$ sits where $\alpha$ crosses zero. Line the three graphs up vertically and those crossings should align.
A sign change in $\omega$ means the system reversed. A sign change in $\alpha$ with $\omega$ keeping its sign means the system is still turning the same way and merely stopped speeding up.

Traps, and the units question that decides half of them

Radians are required and radians are unitless. Statement 5.1.A.1 defines angular displacement in radians. The radian is a ratio of an arc length to a radius, so it carries no dimension, which is why $v = r\omega$ produces metres per second with no conversion factor. Feed degrees or revolutions into any equation on this page and the answer is wrong by a factor of $\pi/180$ or $2\pi$ . Convert first, every time. The radian glossary entry has the conversions.

Average and instantaneous are different quantities in this course. AP Physics 1's Topic 5.1 defines only the averages, so the distinction has nowhere to bite there. Here it does, and worked example 1 is built on a case where the average over an interval is nonzero while the instantaneous value at the end of it is zero.

Constant angular acceleration is a hypothesis, not a default. Check it before reaching for 5.1.A.4.i.

A rigid system is not an object. Statement 5.1.A.1.i says so directly, because different points on it move in different directions. Statement 5.1.A.1.iii gives the one licence to treat it as an object: when the rotation about its own axis can be neglected relative to the motion you care about, as with Earth orbiting the Sun.

Declare the positive sense of rotation and hold it. Statement 5.1.A.1.ii puts the choice in your hands and says nothing about which way to choose. An axis that flips halfway through a solution is a defect even when every individual line is right.

Zero angular velocity does not mean zero angular acceleration. At the two instants in worked example 1 where the rotor is momentarily not turning, its angular acceleration has magnitude $6.0\ \mathrm{rad/s^2}$ . This is the rotational copy of a ball at the top of its flight, and it is tested the same way.

If you are taking AP Physics 1, this is not your page

AP Physics C: Mechanics is a calculus-based introductory college-level physics course, equivalent to the first course in an introductory college sequence in calculus-based physics. Its only stated prerequisite is that students should have taken, or be concurrently taking, calculus. AP Physics 1 is algebra-based, and its Topic 5.1 stops at averages over intervals on purpose.

If you are in AP Physics 1, read the [AP Physics 1 Topic 5.1 page](/ap-physics-1/unit-5-torque-and-rotational-dynamics/5-1-rotational-kinematics) instead. It is written to your framework and it stops where your framework stops. This page is for AP Physics C: Mechanics students. Neither page is a simplified version of the other; the shared title is a College Board naming decision.

From here, Topic 5.2 turns these three angular quantities into the linear motion of a point, and it is the one topic in this unit whose required content is identical in the two courses. The Topic 5.2 page says so plainly and spends its length on what the C exam does differently with it. For the procedure rather than the framework, the rotational kinematics guide owns the step-by-step routine, and the Unit 5 hub lists all six topics.

A rotor whose angular acceleration is not constant

A turbine rotor's angular position about its axis is $\theta(t) = (2.0\ \mathrm{rad/s^3})t^3 - (9.0\ \mathrm{rad/s^2})t^2 + (12\ \mathrm{rad/s})t$ , with $t$ in seconds and counterclockwise taken as positive. Find (a) $\omega(t)$ and $\alpha(t)$ , (b) every time in the first three seconds at which the rotor is momentarily not turning, (c) the angular acceleration at each of those times, and (d) the rotor's average angular velocity over the interval from $t = 0$ to $t = 2.0$ s.

Declare the convention before anything else: counterclockwise is positive, so a negative $\omega$ means the rotor is turning clockwise. Statement 5.1.A.1.ii leaves this choice to you, and it has to hold to the last line.
(a) Apply 5.1.A.2. Differentiating term by term with the power rule, $\omega = \frac{d\theta}{dt} = (6.0)t^2 - (18)t + 12$ , in $\mathrm{rad/s}$ .
Apply 5.1.A.3. Differentiating again, $\alpha = \frac{d\omega}{dt} = (12)t - 18$ , in $\mathrm{rad/s^2}$ . Note that $\alpha$ depends on $t$ , so none of the three equations in 5.1.A.4.i applies anywhere in this problem.
(b) Set $\omega = 0$ . Factoring, $6(t^2 - 3t + 2) = 6(t-1)(t-2) = 0$ , so $t = 1.0$ s and $t = 2.0$ s. Both lie in the first three seconds.
(c) At $t = 1.0$ s, $\alpha = 12(1.0) - 18 = -6.0\ \mathrm{rad/s^2}$ . At $t = 2.0$ s, $\alpha = 12(2.0) - 18 = +6.0\ \mathrm{rad/s^2}$ . The rotor is momentarily at rest at both instants and its angular acceleration is nonzero at both, with opposite signs.
Read what that means physically. Between $t = 1.0$ s and $t = 2.0$ s the sign of $\omega$ is negative, so the rotor turns clockwise through that whole second before reversing again. Check it: $\theta(1.0) = 2.0 - 9.0 + 12 = 5.0$ rad and $\theta(2.0) = 16 - 36 + 24 = 4.0$ rad, so it gives back exactly 1.0 rad.
(d) The average angular velocity over $0$ to $2.0$ s is $\Delta\theta/\Delta t = (4.0 - 0)/2.0 = 2.0\ \mathrm{rad/s}$ , since $\theta(0) = 0$ .
Compare (b) and (d). The average over the interval is $2.0\ \mathrm{rad/s}$ and the instantaneous value at the end of the interval is exactly zero. In AP Physics 1 the framework names only the average, so the two cannot be told apart; here they are different quantities and the exam can ask for either.

(a) $\omega = 6.0t^2 - 18t + 12$ and $\alpha = 12t - 18$ . (b) $t = 1.0$ s and $t = 2.0$ s. (c) $-6.0\ \mathrm{rad/s^2}$ and $+6.0\ \mathrm{rad/s^2}$ : momentarily at rest with nonzero angular acceleration both times. (d) $2.0\ \mathrm{rad/s}$ , which is not the instantaneous value at either end of the interval.

Integrating a non-constant angular acceleration, and what the constant-alpha kit would have given

A wheel starts from rest at $\theta_0 = 0$ and is driven with angular acceleration $\alpha(t) = (0.60\ \mathrm{rad/s^4})t^2$ . Find (a) $\omega(t)$ and $\theta(t)$ , (b) the angular velocity and angular displacement at $t = 5.0$ s, and (c) what $\theta = \frac{1}{2}\alpha t^2$ and $\omega^2 = 2\alpha\theta$ would give if you substituted the angular acceleration at $t = 5.0$ s into them.

(a) Turn 5.1.A.3 around. Since $\alpha = d\omega/dt$ , integrating gives $\omega(t) = \omega_0 + \int_0^t \alpha(t')\, dt'$ . The wheel starts from rest, so $\omega_0 = 0$ .
$\omega(t) = \int_0^t (0.60)t'^2\, dt' = (0.60)\frac{t^3}{3} = (0.20\ \mathrm{rad/s^4})t^3$ . The power rule and the integral of $x^n$ are both printed in the sheet's calculus box.
Turn 5.1.A.2 around the same way: $\theta(t) = \theta_0 + \int_0^t \omega(t')\, dt' = \int_0^t (0.20)t'^3\, dt' = (0.050\ \mathrm{rad/s^4})t^4$ .
(b) At $t = 5.0$ s, $\omega = (0.20)(5.0)^3 = (0.20)(125) = 25\ \mathrm{rad/s}$ and $\theta = (0.050)(5.0)^4 = (0.050)(625) = 31.25 \approx 31$ rad.
(c) The instantaneous angular acceleration at that moment is $\alpha(5.0) = (0.60)(25) = 15\ \mathrm{rad/s^2}$ . Substituting it into $\theta = \frac{1}{2}\alpha t^2$ gives $\frac{1}{2}(15)(25) = 187.5$ rad, six times the true 31.25 rad.
Substituting it into $\omega^2 = 2\alpha\theta$ with the true $\theta = 31.25$ rad gives $\omega = \sqrt{2(15)(31.25)} = \sqrt{937.5} = 30.6\ \mathrm{rad/s}$ against the true $25\ \mathrm{rad/s}$ , about 22% high.
Both failures come from the same place. The derivation of 5.1.A.4.i pulls $\alpha$ out of the integral, which is only allowed when $\alpha$ does not depend on $t$ . Statement 5.1.A.4.i states that condition in its own opening clause.
Sanity check the integration instead: differentiating $\theta = 0.050t^4$ gives $0.20t^3$ , which is $\omega(t)$ , and differentiating again gives $0.60t^2$ , which is the $\alpha(t)$ you started from. Two derivatives should always return you to the given.

(a) $\omega(t) = (0.20\ \mathrm{rad/s^4})t^3$ and $\theta(t) = (0.050\ \mathrm{rad/s^4})t^4$ . (b) $\omega = 25\ \mathrm{rad/s}$ and $\theta = 31$ rad. (c) The constant-angular-acceleration equations give 187.5 rad and $30.6\ \mathrm{rad/s}$ , both wrong, because $\alpha$ is not constant here.

A centrifuge slowing at constant angular acceleration, worked symbolically first

A centrifuge rotor spinning at 1200 revolutions per minute is switched off and comes to rest in 45 s at constant angular acceleration. Find (a) its initial angular velocity in radians per second, (b) its angular acceleration, and (c) the number of revolutions it turns through while stopping. Give the answer to (c) symbolically before putting numbers in.

(a) Convert to radians first, because every equation here assumes them. $\omega_0 = 1200\ \frac{\text{rev}}{\text{min}} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{s}} = 40\pi = 125.66\ \mathrm{rad/s}$ .
Check that the constant-angular-acceleration kit is legal here. The problem states the angular acceleration is constant, so 5.1.A.4.i applies. This is the case AP Physics 1 restricts itself to.
(b) From $\omega = \omega_0 + \alpha t$ with $\omega = 0$ : $\alpha = -\omega_0/t = -40\pi/45 = -8\pi/9 = -2.79\ \mathrm{rad/s^2}$ . The sign is negative because the rotor slows while turning in the positive sense.
(c) Symbolically, with $\omega = 0$ the second equation reduces to $\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ , and substituting $\alpha = -\omega_0/t$ collapses it to $\Delta\theta = \frac{1}{2}\omega_0 t$ . The number of revolutions is $N = \Delta\theta/2\pi = \omega_0 t/4\pi$ .
Substitute: $N = (40\pi)(45)/(4\pi) = 1800\pi/4\pi = 450$ revolutions. The $\pi$ cancels exactly, which is a sign the conversion was set up right.
Cross-check with the third equation, which does not involve $t$ : $0 = \omega_0^2 + 2\alpha\Delta\theta$ gives $\Delta\theta = -\omega_0^2/2\alpha = -(40\pi)^2/(2 \times -8\pi/9) = 1600\pi^2 \times 9/(16\pi) = 900\pi$ rad, and $900\pi/2\pi = 450$ revolutions. The two routes agree.
A quick plausibility check without any algebra: the rotor averages half of 1200 rev/min while it slows uniformly, which is 600 rev/min, and 45 s is three quarters of a minute. $600 \times 0.75 = 450$ .

(a) $\omega_0 = 40\pi = 126\ \mathrm{rad/s}$ . (b) $\alpha = -8\pi/9 = -2.79\ \mathrm{rad/s^2}$ . (c) $N = \omega_0 t/4\pi = 450$ revolutions, confirmed by two independent equations and by an average-rate estimate.

Frequently asked questions

What is the difference between rotational kinematics in AP Physics C and AP Physics 1?

The definitions of angular velocity and angular acceleration. AP Physics 1's essential knowledge 5.1.A.2 and 5.1.A.3 define the average angular velocity as the change in angular position divided by the change in time and the average angular acceleration as the change in angular velocity divided by the change in time. AP Physics C: Mechanics prints the same two statements with the word average removed and with derivatives in place of the difference quotients: angular velocity is the derivative of angular position with respect to time, and angular acceleration is the derivative of angular velocity with respect to time. Both courses print the same three constant-angular-acceleration equations and the same definition of angular displacement. The practical consequence is that a Physics C question can give you angular position as a function of time and ask for an instantaneous value, which the algebra-based framework has no way to pose.

Are the rotational kinematic equations on the AP Physics C Mechanics equation sheet?

Yes, all three, in the rotational column of the mechanics table: angular velocity equals initial angular velocity plus angular acceleration times time; angular position equals initial angular position plus initial angular velocity times time plus one half angular acceleration times time squared; and angular velocity squared equals initial angular velocity squared plus twice the angular acceleration times the change in angular position. The sheet also prints the two derivative definitions above them, which the AP Physics 1 sheet does not have. One Topic 5.1 equation is not printed: the definition of angular displacement as the final angle minus the initial angle. The sheet also omits the angular integral forms even though it prints both linear ones, so if you need the change in angular velocity as the integral of angular acceleration you write that line yourself.

When can I not use the constant angular acceleration equations?

Whenever the angular acceleration changes with time, which essential knowledge 5.1.A.4.i rules out in its opening words: it introduces the three equations with the condition "for constant angular acceleration". Three common signals that the condition fails are an angular position or angular velocity given as a function of time with a cubic or higher power in it, a torque that varies with time or with angle, and a graph of angular acceleration against time that is not a horizontal line. In those cases go back to the definitions and integrate: the change in angular velocity is the integral of angular acceleration over time, and the change in angular position is the integral of angular velocity over time. Substituting a single instantaneous value of angular acceleration into the constant-acceleration equations does not approximate the right answer; for an angular acceleration proportional to time squared it overstates the angle by a factor of six.

Is the direction of angular velocity tested on the AP Physics C Mechanics exam?

No. The Topic 5.1 boundary statement says AP Physics C: Mechanics expects students to be able to mathematically manipulate the magnitudes of angular displacement, angular velocity, and angular acceleration using vector conventions, and then states that the directions of those vectors will not be assessed on the exam. A second paragraph limits descriptions of the directions of rotational kinematics quantities for a point or rigid body to clockwise and counterclockwise with respect to a given axis of rotation. Note that this restriction applies to the kinematic quantities only. Topic 5.3 in the same unit prints no boundary statement and does teach the direction of the torque vector, including the right-hand rule, so the unit treats torque directions and angular-velocity directions differently on purpose.

Why do angular quantities have to be in radians?

Because the radian is defined as the ratio of an arc length to a radius, which makes it dimensionless, and every relation that converts between angular and linear quantities relies on that. Essential knowledge 5.1.A.1 defines angular displacement as the measurement of the angle in radians through which a point on a rigid system rotates about a specified axis. If you feed degrees into the arc-length relation or into the relation between linear speed and angular velocity, the result is wrong by a factor of 180 divided by pi, and revolutions are wrong by a factor of 2 pi. Convert to radians as the first line of any solution: one revolution is 2 pi radians, and one radian is about 57.3 degrees.

How do I find angular velocity from an angular position versus time graph?

Take the slope at the instant you care about. Angular velocity is defined in AP Physics C: Mechanics as the derivative of angular position with respect to time, so the instantaneous angular velocity is the slope of the tangent line to the angular position graph at that moment, not the slope of a chord between two points. Reading a chord gives you the average angular velocity over the interval, which is a different quantity in this course. The same rule one level down gives angular acceleration as the slope of an angular velocity graph. Going the other way, the area under an angular velocity graph between two times is the change in angular position, and the area under an angular acceleration graph is the change in angular velocity, with area below the horizontal axis counting as negative. Essential knowledge 5.1.A.4.ii is the statement that licenses all of this.

Can a rotating system have zero angular velocity and nonzero angular acceleration?

Yes, and it is a standard exam setup. Angular velocity and angular acceleration are independent at any single instant: one is the derivative of angular position and the other is the derivative of angular velocity, and a function can be zero at a point where its derivative is not. A rotor described by a cubic angular position function can pass through angular velocity zero twice, reversing each time, with a nonzero angular acceleration at both instants. The rotational analogue is a ball at the top of its flight, which has zero velocity and the full acceleration due to gravity. In dynamics terms, essential knowledge 5.6.A.1 says angular velocity changes whenever the net torque is not zero, and a system momentarily at rest can certainly have a net torque on it.